Question Video: Analyzing the End Behavior of a Function from Its Graph and Definition Mathematics • 10th Grade

Consider the following graph of 𝑦 = βˆ’(3/(π‘₯ βˆ’ 2)). By looking at the graph and substituting a few successively larger values of π‘₯ into the function, what is the end behavior of the graph as π‘₯ increases along the positive π‘₯-axis?

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Video Transcript

Consider the following graph of 𝑦 equals negative three over π‘₯ minus two. By looking at the graph and substituting a few successively larger values of π‘₯ into the function, what is the end behavior of the graph as π‘₯ increases along the positive π‘₯-axis?

In this question, we’re given the graph of a rational function 𝑦 equals negative three over π‘₯ minus two and asked to use this graph and the function to determine the end behavior of its graph. To do this, we can first recall that the end behavior of a graph is the behavior of the graph as the values of π‘₯ increase or decrease without bound. We want to determine the end behavior of the graph as the values of π‘₯ increase without bounds. And we are told to analyze the end behavior using two different methods. First, we are told to use the given graph to analyze the end behavior. Second, we are told to analyze the end behavior by substituting successively larger values of π‘₯ into the given function.

Let’s start by looking at the given graph. We recall that the π‘₯-coordinate of a point on the graph represents the input value of the function and the 𝑦-coordinate represents the corresponding output of the function. From the graph, we can see that as our values of π‘₯ get larger and larger, the graph of the function approaches the 𝑦-axis from below. This appears to be a horizontal asymptote at 𝑦 equals zero. Since the graph approaches the line 𝑦 equals zero as the values of π‘₯ increase without bound, this implies that the value of 𝑦 approaches zero as π‘₯ increases.

This is enough to answer the question. However, we are asked to check this answer by substituting successively larger input values into the function. To do this, let’s first look at the rational function. We note that the denominator is π‘₯ minus two. This means that we can make calculations easier by making the denominator easier to evaluate. For example, if we substitute π‘₯ equals 12 into the function, then the denominator will evaluate to give 10, and division by 10 is easy to calculate. We have that when π‘₯ equals 12, 𝑦 is equal to negative three divided by 12 minus two, which we can evaluate to give negative 0.3.

We can follow the same process with a larger input value. This time we will use π‘₯ equals 102. This time, the denominator evaluates to give 100, and we can calculate that when π‘₯ is equal to 102, 𝑦 is equal to negative 0.03. This pattern continues indefinitely. For example, we can calculate that when π‘₯ is 1,002, 𝑦 is equal to negative 0.003. We see that as we are increasing the input values of π‘₯, the magnitude of the output values of 𝑦 are getting smaller and smaller. In fact, they are approaching the value zero from the negative direction. This agrees with our analysis from the graph. So we can conclude that as the values of π‘₯ increase along the positive π‘₯-axis without bound, the values of 𝑦 will approach zero.

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